# Thread: Range and inverse of bijective functions

1. Originally Posted by Raoh
Yes the same steps
$f(x)=(x+1)^2\Rightarrow y=(x+1)^2\Leftrightarrow \sqrt{y}=x+1\Leftrightarrow x=\sqrt{y}-1.$
therefore $f^{-1}(y)=\sqrt{y}-1.$
I'm trying

Range = (-2, inf)?

2. Originally Posted by Hellbent
Range = (-2, inf)?
If $f: A \to B$, then $f^{-1}: B \to A$, provided the inverse exists. Look at what you defined your domain, or $A$, to be in the initial problem.

3. Originally Posted by rowe
If $f: A \to B$, then $f^{-1}: B \to A$, provided the inverse exists. Look at what you defined your domain, or $A$, to be in the initial problem.
The domain is any number that's ≤ -2.

4. Originally Posted by Hellbent
The domain is any number that's ≤ -2.
Right, so the range of the inverse cannot be $(-2, \infty)$.

5. Originally Posted by rowe
Right, so the range of the inverse cannot be $(-2, \infty)$.
I know that the range is the set of output numbers, but I'm not understanding it here.
(1, inf)?

6. Originally Posted by Hellbent
I know that the range is the set of output numbers, but I'm not understanding it here.
(1, inf)?
In your above problem, you have given:

f : x |→ (x + 1)² (x ∈ R, x ≤ -2)?

So, we know the domain of $f$ to be $(-\infty, -2]$. We know also know that if $f: A \to B$, then $f^{-1}:B \to A$, and we can therefore conclude that the range of $f^{-1}$ is $(-\infty, -2]$.

7. THE DOMAIN OF $f$IS THE RANGE OF $f^{-1}$ AND VICE VERSA.

8. Thanks, rowe

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